A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20^ C are 440 mm Hg for pentane and 120 ~mm Hg for hexane. the mole fraction of pentane in the vapour phase would

Correct answer: D. 0.478 aligned n_ C_5 H_ 12 n_ C_6 H_ 14 & =(1)/(4) X_ C_3 H_ 12 & =(1)/(5) and X_ C_6 H_ 14 =(4)/(5) P_ C_5 H_ 12 ^ ^ & =440 mm Hg; P_ C_6 H_ 14 ^ ^ & =120 mm Hg P_ T & =P_ C_5 H_ 12 ^ ^ X_ C_5 H_ 12 & +P_ C_6 H_ 14 ^ ^ X_ C_6 H_ 14 & =440 × (1)/(5)+120 × (4)/(5) & =184 mm of Hg aligned By Raoult's Law, P_ C_5 H_ 12 =P_ C_5 H_ 12 ^ ^ X_ C_5 H_ 12 By Dalton's Law, P_ C_5 H_ 12 =X_ C_5 H_ 12 ^ P From (1) and (2), aligned P_ C_5 H_ 12 & =440 × (1)/(5)=88 mm of Hg 88 & =X_ C_5 H_ 12 ^ ×184 X^ & =(88)/(184); X^ =0.478 . aligned

NEET 2005

Exam: NEET
Year: 2005
Subject: Chemistry
Chapter: Solutions

NEET 2005 · Chemistry · Solutions

A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour...

A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at

20^{\circ} \mathrm{C}

are 440

\mathrm{mm} \mathrm{Hg}

for pentane and

120 \mathrm{~mm} \mathrm{Hg}

for hexane. the mole fraction of pentane in the vapour phase would be:

Options

A0.200
B0.549
C0.786
D0.478

Correct answer

D. 0.478

Step-by-step solution

\begin{aligned}\frac{n_{\mathrm{C}_5 \mathrm{H}_{12}}}{n_{\mathrm{C}_6 \mathrm{H}_{14}}}& =\frac{1}{4}\\ \Rightarrow \mathrm{X}_{\mathrm{C}_3 \mathrm{H}_{12}}& =\frac{1}{5} \text{ and } \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}}=\frac{4}{5}\\ \mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{{}^{\circ}}& =440 \mathrm{ mm} \mathrm{Hg}; \\ \mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{{}^{\circ}}& =120 \mathrm{ mm} \mathrm{Hg}\\ \mathrm{P}_{\mathrm{T}}& =\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{{}^{\circ}} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}\\ & +\mathrm{P}_{\mathrm{C}_6 \mathrm{H}_{14}}^{{}^{\circ}} \mathrm{X}_{\mathrm{C}_6 \mathrm{H}_{14}}\\ & =440 \times \frac{1}{5}+120 \times \frac{4}{5}\\ & =184 \mathrm{ mm} \text{ of } \mathrm{Hg} \end{aligned}

By Raoult's Law,

\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}^{{}^{\circ}} \mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}

By Dalton's Law,

\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}=\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \mathrm{P}

From (1) and (2),

\begin{aligned}\mathrm{P}_{\mathrm{C}_5 \mathrm{H}_{12}}& =440 \times \frac{1}{5}=88 \mathrm{ mm} \text{ of } \mathrm{Hg}\\ \Rightarrow \quad 88 & =\mathrm{X}_{\mathrm{C}_5 \mathrm{H}_{12}}^{\prime} \times184 \\ \mathrm{X}^{\prime} & =\frac{88}{184}; \mathrm{X}^{\prime}=0.478 . \end{aligned}

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